Chapter 3.4, Problem 8QR

To determine

To find: The interval where $\frac{dy}{dx}>0$ by analysing the equation algebraically.

Answer to Problem 8QR

The interval where $\frac{dy}{dx}>0$ is $\left(-\infty ,5\right]$.

Explanation of Solution

Given Information: The equation of the quadratic function is $f\left(x\right)=-16x^2+160x-256$

Calculation:

Consider the equation $f\left(x\right)=-16x^2+160x-256$.

The derivative of the function is,

  $\begin{array}{c}y=-16x^2+160x-256\\ \frac{dy}{dx}=\frac{d}{dx}\left(-16x^2+160x-256\right)\\ =-32x+160\end{array}$

The interval where $\frac{dy}{dx}>0$ is,

  $\begin{array}{c}\frac{dy}{dx}>0\\ -32x+160>0\left(\text{Substitute }\frac{dy}{dx}=-32x+160\right)\\ -32x>-160\left(\text{Subtract 160 from both sides}\right)\\ x<\frac{-160}{-32}\left(\text{Divide both sides by -32}\right)\\ =5\left(\text{Simplify}\right)\end{array}$

Therefore, for all values less than 5, $\frac{dy}{dx}>0$.

That is, the interval where $\frac{dy}{dx}>0$ is $\left(-\infty ,5\right]$

Conclusion:

The interval where $\frac{dy}{dx}>0$ is $\left(-\infty ,5\right]$